Cantor Confusion
in [Math]
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From: WM on 23 May 2007 10:46 On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > > According to current mathematics, pi is well defined. Even according > > > > > > to MatheRealism pi is well defined (as an idea). > > > > > > > > > > Your distinction between "number" and "idea" is just terminology, and not > > > > > more than that. > > > > > > > > Its is more. You cannot answer the question whether the numbers P = > > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 nsatisfy P > > > > < P'. > > > > > > Yes, so what? Your distinction is just terminology, and not more than > > > that. > > > > It is by far more than a difference in terminology if one will never > > or always be able to answer a question. > > Makes absolutely no sense. You have some implicit notion of number in mind > that I do not have. Trichotomy. > > You are obviously confused. For an injection from paths to nodes you need > a catalogue of all paths. The tree *is* a catalogue of path, namely of *all* existing paths which represent existing real numbers of the interval [0, 1]. Each of these paths p can only be distinguished from another path p' when both have been separated from each other. You know that each node separates one more path, the number of separated paths minus number of nodes is 2-1-1+2-1-1+2-1-1 +-... Therefore, there must be as many separation points, or nodes, as separated paths. > > > > > Map every node onto the path which leaves it to the left-hand side. > > > > > > I would think that that is *not* an injection from paths to nodes. > > > Moreover, at each node there are many paths that leave it on the left-hand > > > side, so it is not even an injection. > > > > Here is another mapping: Map the node on one of the many paths, call > > it A. The others will get their nodes when they separate from A. > > Again, you are mapping nodes to paths. Why can't you read? Nodes can be bijected to paths. Therefore the direction is unimportant. > > > > You do not find any path. You find at most path bunches. > > As each path bunch starts at the root, I find that at each node the > number of path bunches does not increase but diminish. Interesting. > And as at each > node there are many paths to be found that go from that node, I do not > understand how you come at the idea that I do not find any path. > > > > > The number of paths cannot be larger than the number of edges. > > > > > > Why not? > > > > Because every path contains more edges than it has in common with any > > other path. > > That is not a proof. No? And was so much covinced ... Why do you think this is not a proof? But don't come up with male and female dancers. Regards, WM
From: WM on 23 May 2007 11:07 On 22 Mai, 04:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179750646.698997.275...(a)b40g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 21 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > The writers sum the *sequence* 1, 2, 3, ... with a definite and unique > > > > result (no definition could lead to another result. Your claim that > > > > this was not done or was impossible is wrong. > > > > > > No. They do not state it is a sequence. > > > > They write 1 + 2 + 3 + .... That is the sum of a sequence, i.e., a > > series. > > Yes, you like to use their informal statements as formal statements. 1) Even informal statements are not always wrong. 2)This equation is a formal statement. 3)Is their theorem formal enough? 1.3 Theorem Let L be an infinite cardinal, let K_A (A < L) be nonzero cardinal numbers, and let K = sup {K_A | (A < L)}. Then Sum {A < L} K_A = L * sup {K_A | (A < L)} > > > > But definitions *can* lead to > > > another result. > > > > For instance? > > Definition: sum{i in N} i = 0. Wrong definition. > > > > > > I said that the sequence of natural numbers can be summed up. This is > > > > true. > > > > > > Depends. Did you miss the point that the axiom of choice was required? > > > > It is required for many tasks. > > Yes, so you can *not* state that the set of natural numbers can be summed up. If I can state that the union of a countable collection of countable sets is countable, then I can state that the set of natural numbers can be summed up. When I used the countable union of countable sets nobody opposed up to now. > > > But > > > *their* sum is not a sum of natural numbers, but of cardinal numbers, and > > > it is not a sequence. > > > > Their notation shows the sum of the sequence of natural numbers > > > > 1 + 2 + 3 + ... + n + ... (n in N) > > > > They write, it is easy to see that this sum is equal to aleph_0. > > And they also write that they are doing cardinal arithmetic. All natural numbers are cardinal numbers! > > > > They do only show it when you use *their* definitions (and when you > > > consider cardinal arithmetic). > > > > It is a common and natural definition. That is proven by the fact that > > there is no other result possible. > > But there are other results possible. Consider my definition: > sum{i in N} = 0 > now try to prove that that is not possible. Wrong definition. Proof. 1 > 0. There is no negative natural number. Every sum of natural numbers is >= 1 > 0. > > > Your definition would prove aleph_0 in N. > > Not at all. > > > And n fact, > > if there were all the natural numbers actually existing, then an > > infinite one must be among them. > > A common assertion by you, unproven. Pray give a proof, assuming the > axiom of infinity, and with using the negation of that axiom. 1+2+3+...+X = X(X+1)/2 for every X in N. Proof by induction left as an exercise. Regards, WM
From: WM on 23 May 2007 11:14 On 22 Mai, 04:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179750984.210277.71...(a)n15g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 21 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Pardon, I meant "countably many definitions". This is implied by the > > > > finity of every definition. If there were infinite definitions, then > > > > there were uncountably many definitions. > > > > > > Yup. So the question remains: "you disagree with common mathematical > > > terminology?" > > > > No, it is obvious is that a finite definition means a definition which > > is given by a finite number of words. Infinite definitions are > > nonsense, according to Cantor. > > Yup, you are still living some hundred years ago. Mathematics has gone > forward since that time. Unfortunately today "finite" is meaning "infinite" only when "forward" is meaning "backwards". > > > > But whatever, you can not apply the diagonal argument > > > to "finitely defined numbers". You can supply a list of finite > > > definitions, but not all of them define a number. And for the > > > diagonal argument a list of numbers is needed. > > > > Correct. Therefore it is clear that the finitely defined numbers > > cannot be put in a list, i.e., they cannot be in bijection wit N. > > They are in bijection with a subset of N, so see below. > > > My > > argument should only show that there are countable sets (like the > > paths in the tree) which cannot be put in a bijection with N. > > They can, but that requires an infinite definition. That means they cannot. By "can" I understand: "It can be done in real world on a sheet of paper, a blackboard or in a brain." I do not understand "cannot" by "can", even if this is not up to date. Regards, WM
From: Virgil on 23 May 2007 16:02 In article <1179931587.894431.146050(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM > > > > <mueck...(a)rz.fh-augsburg.de> writes: > > > > > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > > > > According to current mathematics, pi is well defined. Even > > > > > > > according > > > > > > > to MatheRealism pi is well defined (as an idea). > > > > > > > > > > > > Your distinction between "number" and "idea" is just terminology, > > > > > > and not > > > > > > more than that. > > > > > > > > > > Its is more. You cannot answer the question whether the numbers P = > > > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 > > > > > nsatisfy P > > > > > < P'. > > > > > > > > Yes, so what? Your distinction is just terminology, and not more than > > > > that. > > > > > > It is by far more than a difference in terminology if one will never > > > or always be able to answer a question. > > > > Makes absolutely no sense. You have some implicit notion of number in mind > > that I do not have. > > Trichotomy. > > > > You are obviously confused. For an injection from paths to nodes you need > > a catalogue of all paths. > > The tree *is* a catalogue of path, namely of *all* existing paths > which represent existing real numbers of the interval [0, 1]. Each of > these paths p can only be distinguished from another path p' when both > have been separated from each other. This "separation" requirement implies that they are not all separate ab initio, which is false. There is a separate path for each possible subset of the set of levels, and those paths are as separate as are subsets of the set of levels. > > You know that each node separates one more path, the number of > separated paths minus number of nodes is 2-1-1+2-1-1+2-1-1 +-... WM concentrates on the irrelevancies to the exclusion of what is relevant, so as not to see what he does not what to see. No path exists except by being separate from all other paths. This separateness can be fully realized by comparing the sets of levels from which paths branch left, different sets means different paths, and every set determines a path. > > Therefore, there must be as many separation points, or nodes, as > separated paths. Work with sets of levels and you will avoid your present confusion. > > > > > > > Map every node onto the path which leaves it to the left-hand side. > > > > > > > > I would think that that is *not* an injection from paths to nodes. > > > > Moreover, at each node there are many paths that leave it on the > > > > left-hand > > > > side, so it is not even an injection. > > > > > > Here is another mapping: Map the node on one of the many paths, call > > > it A. The others will get their nodes when they separate from A. > > > > Again, you are mapping nodes to paths. Why can't you read? > > Nodes can be bijected to paths. Therefore the direction is > unimportant. Show us any such alleged bijection. While the set of nodes can be injected to the set of paths, I will not believe in a surjection until one is provided. While the set of paths can be surjected to the set of nodes, I will not believe in an injection until one is provided. > > > > > > You do not find any path. You find at most path bunches. > > > > As each path bunch starts at the root, I find that at each node the > > number of path bunches does not increase but diminish. > > Interesting. Makes at least as much sense as WM has been making. > > > And as at each > > node there are many paths to be found that go from that node, I do not > > understand how you come at the idea that I do not find any path. > > > > > > > The number of paths cannot be larger than the number of edges. > > > > > > > > Why not? > > > > > > Because every path contains more edges than it has in common with any > > > other path. > > > > That is not a proof. > > No? And was so much covinced ... Why do you think this is not a proof? Because (1) in a CIBT it is false, and (2) You have shown no reason why what holds for finite trees in that respect, holds for infinite trees > But don't come up with male and female dancers. Why not? They are as relevant as most of WM's reasonings.
From: Virgil on 23 May 2007 16:20
In article <1179932826.710774.250710(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Mai, 04:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179750646.698997.275...(a)b40g2000prd.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > The writers sum the *sequence* 1, 2, 3, ... with a definite and > > > > > unique > > > > > result (no definition could lead to another result. Your claim that > > > > > this was not done or was impossible is wrong. > > > > > > > > No. They do not state it is a sequence. > > > > > > They write 1 + 2 + 3 + .... That is the sum of a sequence, i.e., a > > > series. > > > > Yes, you like to use their informal statements as formal statements. > > 1) Even informal statements are not always wrong. Nor are they always right, and given the choice, the formal statements are a good deal more likely to be right. So anyone consistently choosing the informal may have ulterior motives. > 2)This equation is a formal statement. What equation is that? > 3)Is their theorem formal enough? 1.3 Theorem Let L be an > infinite cardinal, let K_A (A < L) be nonzero cardinal numbers, and > let > K = sup {K_A | (A < L)}. Then Sum {A < L} K_A = L * sup {K_A | (A < > L)} There are some essential definitions deliberately elided here. > > > > > > But definitions *can* lead to > > > > another result. > > > > > > For instance? > > > > Definition: sum{i in N} i = 0. > > Wrong definition. Definitions are evalued by their usefulness and general acceptance, not on "rightness" or "wrongness". Any definition that is grammatically correct is "right", though not necessarily of any value.. > > > > Yes, so you can *not* state that the set of natural numbers can be summed > > up. > > If I can state that the union of a countable collection of countable > sets is countable, then I can state that the set of natural numbers > can be summed up. When I used the countable union of countable sets > nobody opposed up to now. Unless you have given a different definition that the usual one for the sum of two naturals, there is no meaning at all to the sum of infinitely many of them. > > And they also write that they are doing cardinal arithmetic. > All natural numbers are cardinal numbers! And ordinal numbers and, in many set theories, sets, but the operation(s) being performed depend on which view of them one is taking. The operation of addition on naturals does not imply that they are cardinals or ordinals of even sets, so none of those properties are relevant. > > > > > > They do only show it when you use *their* definitions (and when you > > > > consider cardinal arithmetic). > > > > > > It is a common and natural definition. That is proven by the fact that > > > there is no other result possible. > > > > But there are other results possible. Consider my definition: > > sum{i in N} = 0 > > now try to prove that that is not possible. > > Wrong definition. What's wrong with it? > Proof. 1 > 0. There is no negative natural number. Every sum of > natural numbers is >= 1 > 0. The empty sum is zero! And if you are defining sums of arbitrary sets of naturals, that sum is required. And n fact, > > > if there were all the natural numbers actually existing, then an > > > infinite one must be among them. > > > > A common assertion by you, unproven. Pray give a proof, assuming the > > axiom of infinity, and with using the negation of that axiom. > > 1+2+3+...+X = X(X+1)/2 > > for every X in N. Proof by induction left as an exercise. As WM has yet to show that he can provide a valid proof of anything, he is not in a position to require others to do his work for him. So until WM provides entire proofs, his claims remain no more than unproven claims. |